where ##A[x]= x^t A x ##, is the associated quadratic from to the matrix ##A##, where here ##A## is positive definite, of rank ##m## and even. (and I think symmetric?)

So I thought that this meant to solve the quadratic ##A[x]= \vec{x^t} A \vec{x} = n ##, for each ##n##, and the representation number is then given by the number of solutions to this?, subject to ## \vec{x} \in Z^{m} ## ,

no not quite, i needed to check this before i can post my full question, and that my interpretation of what the representation number is correct? (otherwise the question im about to post may not make sense)

where ##A[x]= x^t A x ##, is the associated quadratic from to the matrix ##A##, where here ##A## is positive definite, of rank ##m## and even. (and I think symmetric?)

So I thought that this meant to solve the quadratic ##A[x]= \vec{x^t} A \vec{x} = n ##, for each ##n##, and the representation number is then given by the number of solutions to this?, subject to ## \vec{x} \in Z^{m} ## ,

What is ##Z^{m}## here please? ( z the integer symbol)

Many thanks

Okay so on the attachment of extract from my book, I'm not understanding the comment '##Q_{1}(x,y) ## and ##Q_{2}(x,y) ## yeild the same series since they represent the same integers.'

So as I said above my interpretation of how to compute the ##r(n)## was to :

set ##2 Q(x,y) = A(x,y) = n ## , for each ##n## in turn and count the number of solutions to this for each ##n##.

So looking at ##Q_{0}(x,y)##, should find ##2(x^{2}+xy+6y^2)=0## has one solution (i.e ##(x,y)=0##) , ##2(x^{2}+xy+6y^2)=1## should find 2 solutions and ##2(x^{2}+xy+6y^2)=2,3## has no solutions for ##x \in Z^m ##
Is my understanding correct here?

So then looking at ##Q_1 (x,y)## and ##Q_{2} (x,y) ## which differ only on the sign of the ##xy## term, I don't see how it is obvious that these will have the same number of solutions for ##Q(x,y) = n## for each ##n##?